Homotopical nilpotency of self homotopy equivalence

Given a topological space $X$, ${\rm aut}(X)$ denotes the monoid of the homotopy self equivalences of $X$, that are maps $f: X\rightarrow X$ which admits a homotopy inverse. ${\rm aut}_1(X)$ denotes the path component of the identity map. The homotopical niloptency of ${\rm aut}_1(X)$ as H-space, denoted here ${\rm Hnil} ({\rm aut}_1(X))$ is then the least integer $n$ such that $(n+1)$-th commutator is nullhomotopic.

Question: When $G$ is a connected topological group, it is true that $${\rm Nil}(G)\leq {\rm Hnil} ({\rm aut}_1(G)) ?$$

• According to your definition the number on the right is infinity, so technically the answer is yes. – Fernando Muro Dec 11 '14 at 21:41
• @FernandoMuro : Thanks, but why ${\rm Hnil} ({\rm aut}_1(G))$ is always infinite. The same inequality holds for ${\rm aut}(G)$ due to different justification in example 2.7 Y. F\'elix, G. Lupton, S. B. Smith, \textit{The rational homotopy type of the space of self-equivalences of a fibration}, Homology, Homotopy and Applications, vol. \textbf{12}(2), 2010, 371-400. – MyIsmail Dec 11 '14 at 21:56
• Because no map in aut$_1(G)$ is nullhomotopic. – Fernando Muro Dec 11 '14 at 22:20
• @FernandoMuro: Clear and sharp now, thanks – MyIsmail Dec 11 '14 at 22:33
• @MarkGrant sorry, I understood that the image should be nullhomotopic as a selfmap of G. – Fernando Muro Dec 12 '14 at 8:32